The action of a free relativistic particles can be given by
$$S=\frac{1}{2}\int d\tau \left(e^{-1}(\tau)g_{\mu\nu}(X)X^\mu(\tau)X^\nu(\tau)-e(\tau)m^2\right).$$
If we then make an infinitesimal transformation of the parametrization parameter $\tau$ this would be
$$\tau\to\tau^\prime=\tau-\eta(\tau),$$
for an infinitesimal parameter $\eta(\tau)$.

Of course we can describe the system as it pleases us so we know that $$X^{\mu^\prime}(\tau^\prime)=X^\mu(\tau).$$ From this relation we see that $X^{\mu^\prime}(\tau)$ must be
$$X^{\mu^\prime}(\tau)
\approxeq X^{\mu^\prime}(\tau^\prime+\eta(\tau))
\approxeq X^{\mu^\prime}(\tau^\prime)+\eta(\tau)\frac{d X^{\mu^\prime}(\tau^\prime)}{d\tau^\prime}
\approxeq X^{\mu}(\tau)+\eta(\tau)\frac{d X^{\mu}(\tau)}{d\tau}$$
Which all can be summarized as
$$\delta X^{\mu}(\tau)=X^{\mu^\prime}(\tau)-X^{\mu}(\tau)=\eta(\tau)\frac{d X^{\mu}(\tau)}{d\tau}.$$
Now this is all well I hope. But if one does the same argument for $e(\tau)$ one gets the wrong transformation. It is a scalar function so it has to obey
$$e^\prime(\tau^\prime)=e(\tau).$$ Which would give the same transformation.

The right transformation are written in David Tong's lectures on string theory on page 13, eq. 1.10.
The transformation is
$$\delta e(\tau)=\frac{d}{d\tau}\left(\eta(\tau) e(\tau)\right).$$
Could someone show me how this is done and elaborate a little on how one knows how different object transforms?

III) Classical BV formulation. Let us mention for completeness that the gauge transformation $\delta$ can be encoded as a BRST transformation, cf. e.g. Ref. 2 and this Phys.SE post. Roughly speaking, the Grassmann-even gauge parameter $\eta$ is then replaced by a Grassmann-odd Faddeev-Popov (FP) ghost $C$. (Actually, the gauge parameter $\eta$ will more precisely be replaced with the combination $e^{1-r}C$, where $r\in\mathbb{R}$ is a power, to be more general, cf. eq. (16) below.) To minimize the appearances of time derivatives, instead of using the Lagrangian (5), it becomes a bit simpler to start from the Hamiltonian Lagrangian

which should be compared with eq. (1.10). The BV gauge-fixing fermion $\psi$ can be chosen on the form

$$ \psi ~:=~\int \! \mathrm{d}\tau~\bar{C}\left(\frac{\xi}{2}B +\chi(e) +\epsilon \dot{e}\right), \tag{17} $$
where $\xi,\epsilon\in\mathbb{R}$ are gauge-fixing parameters. Moreover, $\chi(e)=(e\!-\!e_0)\chi^{\prime}$ is a gauge-fixing condition (which we will assume is affine in $e$, so that the derivative $\chi^{\prime}$ is constant). The gauge-fixed Lagrangian becomes

where the $\sim$ symbol means equality up to total time derivative terms.
The physical quantities do not depend on the choice of the gauge-fixing fermion $\psi$, as long as certain rank conditions are met.

V) Classical BFV formulation. We identify $p_e\approx\epsilon B$ with the canonical momentum of the einbein $e$, and we identify the antifield $e^{\ast}\equiv \bar{P}$ with the FP ghost momentum. Introduce an ultra-local Poisson bracket $\{\cdot,\cdot\}_{PB}$ with the following canonical pairs

VI) Dirac bracket. Let us integrate out the two FP momenta $P$ and $\bar{P}$.
Then the BFV Lagrangian (29) becomes the gauge-fixed Lagrangian (18) from Section III. The corresponding two 2nd class constraints

VII) Quantum BV formulation. Eqs. (20), (22) & (34) suggest that we should put $r=0$, so let us do this from now on. Inspired by the BFV-BRST transformations (24), we modify the BV Lagrangian (13) into

$\begingroup$Comments for later: It seems the BRST transformation (16) for $x^{\mu}$ should transform into $p$ rather than $\dot{x}$. (Of course, that's the same on-shell.) This is precisely what is done in eq. (36).$\endgroup$
– Qmechanic♦Jan 15 at 12:56

$\begingroup$Comments for later: The abelianization $r=0$ (changing the gauge algebra) may alter the partition function, even for Hamiltonian BFV.$\endgroup$
– Qmechanic♦Jan 21 at 14:08